Problem: The lifespans of zebras in a particular zoo are normally distributed. The average zebra lives $28.5$ years; the standard deviation is $2$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a zebra living longer than $24.5$ years.
Solution: $28.5$ $26.5$ $30.5$ $24.5$ $32.5$ $22.5$ $34.5$ $95\%$ $2.5\%$ $2.5\%$ We know the lifespans are normally distributed with an average lifespan of $28.5$ years. We know the standard deviation is $2$ years, so one standard deviation below the mean is $26.5$ years and one standard deviation above the mean is $30.5$ years. Two standard deviations below the mean is $24.5$ years and two standard deviations above the mean is $32.5$ years. Three standard deviations below the mean is $22.5$ years and three standard deviations above the mean is $34.5$ years. We are interested in the probability of a zebra living longer than $24.5$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $95\%$ of the zebras will have lifespans within 2 standard deviations of the average lifespan. The remaining $5\%$ of the zebras will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({2.5\%})$ will live less than $24.5$ years and the other half $({2.5\%})$ will live longer than $32.5$ years. The probability of a particular zebra living longer than $24.5$ years is ${95\%} + {2.5\%}$, or $97.5\%$.